a) \(\sqrt{8x+4}-\sqrt{\dfrac{10x+5}{5}}-2=\dfrac{1}{2}-\sqrt{18x+9}\) (ĐK: \(x\ge-\dfrac{1}{2}\))

\(\Leftrightarrow2\sqrt{2x+1}-\sqrt{\dfrac{5\cdot\left(2x+1\right)}{5}}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)

\(\Leftrightarrow2\sqrt{2x+1}-\sqrt{2x+1}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)

\(\Leftrightarrow\sqrt{2x+1}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)

\(\Leftrightarrow\sqrt{2x+1}+3\sqrt{2x+1}=\dfrac{1}{2}+2\)

\(\Leftrightarrow4\sqrt{2x+1}=\dfrac{5}{2}\)

\(\Leftrightarrow\sqrt{2x+1}=\dfrac{5}{8}\)

\(\Leftrightarrow2x+1=\dfrac{25}{64}\)

\(\Leftrightarrow2x=-\dfrac{39}{64}\)

\(\Leftrightarrow x=-\dfrac{39}{128}\)

b) \(\sqrt{4x^2-4x+1}-3=2\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=2+3\)

\(\Leftrightarrow\left|2x-1\right|=5\)

TH1: \(\left|2x-1\right|=2x-1\) khi \(x\ge\dfrac{1}{2}\) 

Pt trở thành:

\(2x-1=5\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\left(tm\right)\)

TH2: \(\left|2x-1\right|=-\left(2x-1\right)\) khi \(x< \dfrac{1}{2}\)

Pt trở thành:

\(-\left(2x-1\right)=5\) (ĐK \(x< \dfrac{1}{2}\)

\(\Leftrightarrow2x-1=-5\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\)

Vậy: ..

c: \(\Leftrightarrow\sqrt{4x^2-4x+5}=5-2x\)

=>5-2x>=0 và 4x^2-4x+5=(5-2x)^2

=>x<=5/2 và 4x^2-4x+5=4x^2-20x+25

=>x<=5/2 và -4x+5=-20x+25

=>x<=5/2 và 16x=20

=>x=5/4(nhận)

d: \(\Leftrightarrow x^2-4\sqrt{x-1}-9x+28=0\)

=>\(x^2-5x-4x+20-4\sqrt{x-1}+8=0\)

=>\(x\left(x-5\right)-4\left(x-5\right)-4\left(\sqrt{x-1}-2\right)=0\)

=>\(\left(x-5\right)\left(x-4\right)-4\left(\sqrt{x-1}-2\right)=0\)

=>\(\left(x-4\right)\cdot\left[\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+2\right)\right]-4\left(\sqrt{x-1}-2\right)=0\)

=>\(\left(\sqrt{x-1}-2\right)\left[\left(x-4\right)\left(\sqrt{x-1}+2\right)-4\right]=0\)

=>căn(x-1)-2=0

=>căn(x-1)=2

=>x-1=4

=>x=5

Gọi O là giao của AC với BD trong mp(ABCD)

Trong mp(SBD), gọi E là giao của SO với DM

\(E\in SO\subset\left(SAC\right)\)

\(E\in DM\subset\left(ADM\right)\)

=>E thuộc (SAC) giao (ADM)

mà \(A\in\left(SAC\right)\cap\left(ADM\right)\)

nên \(\left(SAC\right)\cap\left(ADM\right)=EA\)

a: =6x^2-2x+15x-5-6x^2+18x-21x

=13x+18x-21x-5

=10x-5

b; =x^2+2xy+y^2+x^2-2xy+y^2-2x^2

=2x^2+2y^2-2x^2

=2y^2

c: =9x^2+24x+16+16x^2-8x+1+4x^2-25

=29x^2+16x-8

d: =8x^3+1+8-27x^3

=-19x^3+9

a) \(15+\left(-24\right)\)

\(=15-24\)

\(=-9\)

b) \(\left(-135\right)-\left(-35\right)\)

\(=-135+35\)

\(=-100\)

c) \(3-\left(-15\right)\)

\(=3+15\)

\(=18\)

d) \(\left(45-3756\right)+3756\)

\(=45-3756+3756\)

\(=45\)

e) \(-2023-\left(2022-2023\right)\)

\(=-2023-2022+2023\)

\(=-2022\)

f) \(\left(-17\right)+\left(-23\right)\)

\(=-17-23\)

\(=-40\)

g) \(38-50\)

\(=-12\)

h) \(-6-15\)

\(=-21\)

k) \(-75-\left(-80\right)\)

\(=-75+80\)

\(=5\)

Ta có:

\(P\left(x\right)=-4x^4+8x^2-12x+5\)

\(P\left(x\right)=-4x\left(x^3-2x+3\right)+5\)

Thay \(x^3-2x+3=0\) vào P(x) ta có:

\(P\left(x\right)=-4x\cdot0+5=0=5\)

⇒ Chọn B

P=-4x(x^3-2x+3)+5

=-4x*0+5=5

=>B

a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)

\(=\dfrac{2}{5}+\dfrac{4}{2}\)

\(=\dfrac{2}{5}+2\)

\(=\dfrac{2}{5}+\dfrac{10}{5}\)

\(=\dfrac{12}{5}\)

b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)

\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)

\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)

\(=1-\dfrac{2}{2008}\)

\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)

\(=\dfrac{2006}{2008}\)

\(=\dfrac{1003}{1004}\)

a: =2/5+4/2

=2/5+2

=12/5

b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

\(\left(x+2y\right)\left(x-2y\right)-4y\left(x-y\right)\)

\(=x^2-\left(2y\right)^2-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-\left(2y\right)^2\)

\(=\left(x-2y\right)^2-\left(2y\right)^2\)

\(=\left(x-2y-2y\right)\left(x-2y+2y\right)\)

\(=x\left(x-4y\right)\)

1.

$3x^4-48=3(x^4-16)=3[(x^2)^2-4^2]=3(x^2-4)(x^2+4)$

$=3(x-2)(x+2)(x^2+4)$

2. 

$x^3-4x^2+8x-8=(x^3-8)-(4x^2-8x)$
$=(x-2)(x^2+2x+4)-4x(x-2)=(x-2)(x^2+2x+4-4x)=(x-2)(x^2-2x+4)$

3.

$x^4-27x=x(x^3-27)=x(x^3-3^3)=x(x-3)(x^2+3x+9)$

4.

$x^4-x^3+x^2-1=(x^4-x^3)+(x^2-1)=x^3(x-1)+(x-1)(x+1)$

$=(x-1)(x^3+x+1)$

5.

$x^5+x^3-x^2-1=(x^5+x^3)-(x^2+1)=x^3(x^2+1)-(x^2+1)$
$=(x^2+1)(x^3-1)=(x^2+1)(x-1)(x^2+x+1)$

6.

$x^3+2x^2+2x+1=(x^3+x^2)+(x^2+2x+1)=x^2(x+1)+(x+1)^2$
$=(x+1)(x^2+x+1)$

7.

$x^3+x^2-4x-4=(x^3+x^2)-(4x+4)=x^2(x+1)-4(x+1)$

$=(x+1)(x^2-4)=(x+1)(x-2)(x+2)$

8.

$x^4-2x^3+2x-1=(x^4-2x^3+x^2)-(x^2-2x+1)$

$=(x^2-x)^2-(x-1)^2=x^2(x-1)^2-(x-1)^2=(x-1)^2(x^2-1)=(x-1)^2(x-1)(x+1)$

1/2^2+1/3^2+...+1/8^2>1/2-1/3+1/3-1/4+...+1/8-1/9=1/2-1/9=7/18>6/18=1/3(1)

1/2^2+1/3^2+...+1/8^2<1-1/2+1/2-1/3+...+1/7-1/8=7/8<1(2)

Từ (1), (2) suy ra 1/3<1/2^2+1/3^2+...+1/8^2<1

lạm dụng Latex đi anh =))